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JEE Mains · Physics · STD 12 - 8. Electromagnetic waves

A plane electromagnetic wave of wavelength \(\lambda \) has an intensity \(I.\)  It is propagating along the positive \(Y-\)  direction. The allowed expressions for the electric and magnetic fields are given by

  1. A \(\vec E\, = \,\sqrt {\frac{I}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat i\,;\,\vec B\, = \,\frac{1}{c}E\hat k\)
  2. B \(\vec E\, = \,\sqrt {\frac{I}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, =  - \,\frac{1}{c}E\hat i\)
  3. C \(\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, =  + \frac{1}{c}E\hat i\)
  4. D \(\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y + ct)} \right]\,\hat k\,;\,\vec B\, = \frac{1}{c}E\hat i\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, =  + \frac{1}{c}E\hat i\)

Step-by-step Solution

Detailed explanation

If \(E_0\) is magnitude of electric field then \(\frac{1}{2}\,{\varepsilon _0}{E_0}^2\, \times \,C\, = \,I\, \Rightarrow {\kern 1pt} {E_0}\, = \sqrt {\frac{{2I}}{{C{\varepsilon _0}}}} \) \({B_0} = {\kern 1pt} \frac{{{E_0}}}{C}\,\) Direction of \(\vec E \times \vec B\) will be…
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